Question: Let $y=\ln(5x^2+1)$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{5x^2+1}$ (Choice B) B $\dfrac{10x}{5x^2+1}$ (Choice C) C $\dfrac{10x}{\ln(x)(5x^2+1)}$ (Choice D) D $\dfrac{\ln(x)}{5x^2+1}$
Solution: $\ln(5x^2+1)$ is a logarithmic function, but its argument isn't simply $x$. Therefore, it's a composite function. In other words, suppose $u(x)=5x^2+1$, then $y=\ln\Bigl(u(x)\Bigr)$. $\dfrac{dy}{dx}$ can be found using the following identity: $\dfrac{d}{dx}\left[\ln\Bigl(u(x)\Bigr)\right]=\dfrac{u'(x)}{u(x)}$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\\\\\ &=\dfrac{d}{dx}\ln(5x^2+1) \\\\\\\\ &=\dfrac{d}{dx}\ln\Bigl(u(x)\Bigr)&&\gray{\text{Let }u(x)=5x^2+1} \\\\\\\\ &=\dfrac{u'(x)}{u(x)} \\\\\\\\ &=\dfrac{10x}{5x^2+1}&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{10x}{5x^2+1}$.